Question

When photons of energy 5 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $K_{A}eV$and de Broglie wavelength ${\lambda }_A$ The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 5.30 eV is $K_B=(K_A-1.5).eV$ If the de Broglie wavelength of these photoelectrons is ${\lambda }_B=2{\lambda }_A$ then find $K_A$ in electron volt.

Answer 1

$K_A=\frac{hc}{\lambda }-W_A=5-W_A$

$K_B=5.3-W_B$ It is given $K_B=K_A-1.5$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

$\therefore {\lambda }_B^2=4{\lambda }_A^2$

$\frac{h^2}{2m_e{K}_B}=\frac{4h^2}{2m_e{K}_A}\Rightarrow K_A=4K_B$

$\therefore K_B=4K_B-1.5$

$3K_B=1.5$

$\therefore K_B=0.5eV$

$\therefore K_A=2eV$